01-04 - Palindrome Permutation [Cracking the Coding Interview]

function palindromPermutation(str) {
	const string = str.toLowerCase(),
		mapChars = {}

	let oddCharCount = 0,
		totalCharCount = 0

	for (const char of string) {
		//check only lowercase letters (a-z)
		if (char.charCodeAt(0) > 96 && char.charCodeAt(0) < 123) {
			totalCharCount++
			mapChars[char] = !mapChars[char] ? 1 : mapChars[char] + 1
		}
	}

	for (const char in mapChars) {
		const count = mapChars[char]
		if (count % 2 !== 0) oddCharCount++
		if (oddCharCount > 1) return false
	}

	//Allow one odd char only for strings with character count odd
	if (totalCharCount % 2 === 0 && oddCharCount > 0) return false

	return true
}

/*
n = length of string
Space Complexity: O(n)
Time Complexity: O(n)
*/

//Tests

const testCases = [
	['string', false],
	['strts', true],
	['sssstssss', true],
	['settss', false],
	['Tact Coa', true],
	['Tact boa', false],
]

for (const test of testCases) {
	console.log(palindromPermutation(test[0]) === test[1], test)
}